Use the distributivity of multiplication of rational numbers over their addition to simplify:

Question:

Use the distributivity of multiplication of rational numbers over their addition to simplify:

(i) $\frac{3}{5} \times\left(\frac{35}{24}+\frac{10}{1}\right)$

(ii) $\frac{-5}{4} \times\left(\frac{8}{5}+\frac{16}{5}\right)$

(iii) $\frac{2}{7} \times\left(\frac{7}{16}-\frac{21}{4}\right)$

(iv) $\frac{3}{4} \times\left(\frac{8}{9}-40\right)$

Solution:

(i) $\frac{3}{5} \times\left(\frac{35}{24}+\frac{10}{1}\right)=\frac{3}{5} \times \frac{35}{24}+\frac{3}{5} \times \frac{10}{1}=\frac{7}{8}+\frac{6}{1}=\frac{7+48}{8}=\frac{55}{8}$

(ii) $\frac{-5}{4} \times\left(\frac{8}{5}+\frac{16}{5}\right)=\frac{-5}{4} \times \frac{8}{5}+\frac{-5}{4} \times \frac{16}{5}=\frac{-2}{1}+\frac{-4}{1}=-6$

(iii) $\frac{2}{7} \times\left(\frac{7}{16}-\frac{21}{4}\right)=\frac{2}{7} \times \frac{7}{16}-\frac{2}{7} \times \frac{21}{4}=\frac{1}{8}-\frac{3}{2}=\frac{1-12}{8}=\frac{-11}{8}$

(iv) $\frac{3}{4} \times\left(\frac{8}{9}-40\right)=\frac{3}{4} \times \frac{8}{9}-\frac{3}{4} \times 40=\frac{2}{3}-30=\frac{2-90}{3}=\frac{-88}{3}$